For $(a,b)$, if $m^* ((a,b)) = m^* ( (a,b) \cap E ) + m^*( (a,b) \cap - E)$ then $E$ in $\mathbb{R}$ is measurable

Question

If $m^* ((a,b)) = m^* ( (a,b) \cap E ) + m^*( (a,b) - E)$ for all open intervals $(a,b)$, then $E$ in $\mathbb{R}$ is measurable.

How do I prove this? Totally stuck.

Answer 1

You need to show that $m^*(A \cap E) + m^*(A - E) \le m^*(A)$ for every set $A$. Let $\{I_k\}$ be a cover of set $A$ by open intervals. Use monotonity and countable subadditivity to find $$m^*(A \cap E) + m^*(A - E) \le m^* \left( (\cup I_k) \cap E \right) + m^* \left( (\cup I_k) - E \right) \le \sum \left[ m^*(I_k \cap E) + m^*(I_k - E) \right].$$ The last sum (by hypothesis) is $\sum m^*(I_k).$ What is the definition of $m^*(A)$?